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Canonical form/ 2x^2+6xy+2y^2=5

2x^2+6xy+2y^2=5 canonical form

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        2      2            
-5 + 2*x  + 2*y  + 6*x*y = 0
2x2+6xy+2y2−5=02 x^{2} + 6 x y + 2 y^{2} - 5 = 0 
2*x^2 + 6*x*y + 2*y^2 - 5 = 0
Detail solution
Given line equation of 2-order:
2x2+6xy+2y2−5=02 x^{2} + 6 x y + 2 y^{2} - 5 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=2a_{11} = 2
a12=3a_{12} = 3
a13=0a_{13} = 0
a22=2a_{22} = 2
a23=0a_{23} = 0
a33=−5a_{33} = -5
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣2332∣\Delta = \left|\begin{matrix}2 & 3\\3 & 2\end{matrix}\right|
Δ=−5\Delta = -5
Because
Δ\Delta
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
a11x0+a12y0+a13=0a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0
a12x0+a22y0+a23=0a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0
substitute coefficients
2x0+3y0=02 x_{0} + 3 y_{0} = 0
3x0+2y0=03 x_{0} + 2 y_{0} = 0
then
x0=0x_{0} = 0
y0=0y_{0} = 0
Thus, we have the equation in the coordinate system O'x'y'
a33′+a11x′2+2a12x′y′+a22y′2=0a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0
where
a33′=a13x0+a23y0+a33a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}
or
a33′=−5a'_{33} = -5
a33′=−5a'_{33} = -5
then equation turns into
2x′2+6x′y′+2y′2−5=02 x'^{2} + 6 x' y' + 2 y'^{2} - 5 = 0
Rotate the resulting coordinate system by an angle φ
x′=x~cos⁡(ϕ)−y~sin⁡(ϕ)x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}
y′=x~sin⁡(ϕ)+y~cos⁡(ϕ)y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}
φ - determined from the formula
cot⁡(2ϕ)=a11−a222a12\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}
substitute coefficients
cot⁡(2ϕ)=0\cot{\left(2 \phi \right)} = 0
then
ϕ=π4\phi = \frac{\pi}{4}
sin⁡(2ϕ)=1\sin{\left(2 \phi \right)} = 1
cos⁡(2ϕ)=0\cos{\left(2 \phi \right)} = 0
cos⁡(ϕ)=cos⁡(2ϕ)2+12\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}
sin⁡(ϕ)=1−cos⁡2(ϕ)\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}
cos⁡(ϕ)=22\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}
sin⁡(ϕ)=22\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}
substitute coefficients
x′=2x~2−2y~2x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}
y′=2x~2+2y~2y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}
then the equation turns from
2x′2+6x′y′+2y′2−5=02 x'^{2} + 6 x' y' + 2 y'^{2} - 5 = 0
to
2(2x~2−2y~2)2+6(2x~2−2y~2)(2x~2+2y~2)+2(2x~2+2y~2)2−5=02 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 6 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 2 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 5 = 0
simplify
5x~2−y~2−5=05 \tilde x^{2} - \tilde y^{2} - 5 = 0
−5x~2+y~2+5=0- 5 \tilde x^{2} + \tilde y^{2} + 5 = 0
Given equation is hyperbole
x~21−y~25=1\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{5} = 1
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
e⃗1=(22, 22)\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)
e⃗2=(−22, 22)\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)
Invariants method
Given line equation of 2-order:
2x2+6xy+2y2−5=02 x^{2} + 6 x y + 2 y^{2} - 5 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=2a_{11} = 2
a12=3a_{12} = 3
a13=0a_{13} = 0
a22=2a_{22} = 2
a23=0a_{23} = 0
a33=−5a_{33} = -5
The invariants of the equation when converting coordinates are determinants:
I1=a11+a22I_{1} = a_{11} + a_{22}
     |a11  a12|
I2 = |        |
     |a12  a22|

I3=∣a11a12a13a12a22a23a13a23a33∣I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|
I(λ)=∣a11−λa12a12a22−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|
     |a11  a13|   |a22  a23|
K2 = |        | + |        |
     |a13  a33|   |a23  a33|

substitute coefficients
I1=4I_{1} = 4
     |2  3|
I2 = |    |
     |3  2|

I3=∣23032000−5∣I_{3} = \left|\begin{matrix}2 & 3 & 0\\3 & 2 & 0\\0 & 0 & -5\end{matrix}\right|
I(λ)=∣2−λ332−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 3\\3 & 2 - \lambda\end{matrix}\right|
     |2  0 |   |2  0 |
K2 = |     | + |     |
     |0  -5|   |0  -5|

I1=4I_{1} = 4
I2=−5I_{2} = -5
I3=25I_{3} = 25
I(λ)=λ2−4λ−5I{\left(\lambda \right)} = \lambda^{2} - 4 \lambda - 5
K2=−20K_{2} = -20
Because
I2<0∧I3≠0I_{2} < 0 \wedge I_{3} \neq 0
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
−I1λ+I2+λ2=0- I_{1} \lambda + I_{2} + \lambda^{2} = 0
or
λ2−4λ−5=0\lambda^{2} - 4 \lambda - 5 = 0
λ1=5\lambda_{1} = 5
λ2=−1\lambda_{2} = -1
then the canonical form of the equation will be
x~2λ1+y~2λ2+I3I2=0\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0
or
5x~2−y~2−5=05 \tilde x^{2} - \tilde y^{2} - 5 = 0
x~21−y~25=1\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{5} = 1
- reduced to canonical form
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